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Question Number 46851 by maxmathsup by imad last updated on 01/Nov/18

l e t f (x) = \int_{0}^{2 π} \frac{s i n t}{x + s i n t} d t w i t h x > 1

1) c a l c u l a t e f (x)

2) c a l c u l a t e \int_{0}^{2 π} \frac{s i n t}{{(x + s i n t)}^{2}} d t

3) f i n d t h e v a l u e o f \int_{0}^{2 π} \frac{s i n t}{2 + s i n t} d t a n d \int_{0}^{2 π} \frac{s i n t}{{(2 + s i n t)}^{2}} d t

Commented by maxmathsup by imad last updated on 02/Nov/18

1) w e h a v e f (x) = \int_{0}^{2 π} \frac{x + s i n t - x}{x + s i n t} d t = 2 π - x \int_{0}^{2 π} \frac{d t}{x + s i n t}

c h a n g e m e n t e^{i t} = z g i v e \int_{0}^{2 π} \frac{d t}{x + s i n t} = \int_{∣ z ∣= 1} \frac{1}{x + \frac{z - z^{- 1}}{2 i}} \frac{d z}{i z}

= \int_{∣ z ∣= 1} \frac{2 i d z}{i z (2 i x + z - z^{- 1})} = \int_{∣ z ∣= 1} \frac{2 i d z}{- 2 x z + {i z}^{2} - i} = \int_{∣ z ∣= 1} \frac{2 d z}{2 i x z + z^{2} - 1}

= \int_{∣ z ∣= 1} \frac{2 d z}{z^{2} + 2 i x z - 1} l e t φ (z) = \frac{2}{z^{2} + 2 i x z - 1} p o l e s o f φ ?

Δ^{^{'}} = {(i x)}^{2} + 1 = 1 - x^{2} = {(i \sqrt{x^{2} - 1})}^{2} \Rightarrow z_{1} = - i x + i \sqrt{x^{2} - 1}

z_{2} = - i x - i \sqrt{x^{2} - 1}

∣ z_{1} ∣ - 1 =∣ - x + \sqrt{x^{2} - 1} ∣ - 1 = \sqrt{x^{2} - 1} - x - 1 = \sqrt{x^{2} - 1} - (x + 1)

a n d x^{2} - 1 - {(x + 1)}^{2} = x^{2} - 1 - x^{2} - 2 x - 1 = - 2 x - 2 < 0 \Rightarrow∣ z_{1} ∣< 1

∣ z_{2} ∣ - 1 =∣ x + \sqrt{x^{2} - 1} ∣ - 1 = x - 1 + \sqrt{x^{2} - 1} > 0 b e c a u s e x > 1 \Rightarrow z_{2} i s o u t o f c i r c l e

s o \int_{∣ z ∣= 1} φ (z) d z = 2 i π R e s (φ, z_{1}) b u t φ (z) = \frac{2}{(z - z_{1}) (z - z_{2})} \Rightarrow

R e s (φ, z_{1}) = {l i m}_{z \to z_{1}} (z - z_{1}) φ (z) = \frac{2}{z_{1} - z_{2}} = \frac{2}{2 i \sqrt{x^{2} - 1}} \Rightarrow

\int_{∣ z ∣= 1} φ (z) d z = 2 i π \frac{1}{i \sqrt{x^{2} - 1}} = \frac{2 π}{\sqrt{x^{2} - 1}} \Rightarrow ★ f (x) = \frac{2 π}{\sqrt{x^{2} - 1}} ★ w i t h x > 1

Commented by maxmathsup by imad last updated on 02/Nov/18

2) w e h a v e f^{^{'}} (x) = \int_{0}^{2 π} \frac{\partial}{\partial x} {\frac{s i n t}{x + s i n t}} d t = - \int_{0}^{2 π} \frac{s i n t}{{(x + s i n t)}^{2}} d t \Rightarrow

\int_{0}^{2 π} \frac{s i n t}{{(x + s i n t)}^{2}} d t = - f^{^{'}} (x) b u t f (x) = \frac{2 π}{\sqrt{x^{2} - 1}} = 2 π {(x^{2} - 1)}^{- \frac{1}{2}} \Rightarrow

f^{^{'}} (x) = 2 π (- \frac{1}{2}) (2 x) {(x^{2} - 1)}^{- \frac{3}{2}} = \frac{- 2 π x}{{(x^{2} - 1)}^{\frac{3}{2}}} = \frac{- 2 π x}{(x^{2} - 1) \sqrt{x^{2} - 1}} \Rightarrow

\int_{0}^{2 π} \frac{s i n t}{{(x + s i n t)}^{2}} d t = \frac{2 π x}{(x^{2} - 1) \sqrt{x^{2} - 1}} w i t h x > 1 .

Commented by maxmathsup by imad last updated on 02/Nov/18

3) w e h a v e \int_{0}^{2 π} \frac{s i n t}{2 + s i n t} d t = f (2) = \frac{2 π}{\sqrt{2^{2} - 1}} = \frac{2 π}{\sqrt{3}}

\int_{0}^{2 π} \frac{s i n t}{{(2 + s i n t)}^{2}} d t = f^{^{'}} (2) = \frac{4 π}{(2^{2} - 1) \sqrt{2^{2} - 1}} = \frac{4 π}{3 \sqrt{3}} .